Question

If $2ta{n}^{-1}\left(cosx\right)=ta{n}^{-1}\left(cose{c}^{2}x\right)$, then x =

• A. $\frac{\pi }{2}$
• B. $\pi$
• C. $\frac{\pi }{6}$
• D. $\frac{\pi }{3}$

Answer 1

The correct option is D $\frac{\pi }{3}$

Given, $2{\tan }^{-1}\left(\cos x\right)={\tan }^{-1}\left(cose{c}^{2}x\right)$
$\Rightarrow {\tan }^{-1}\left(\frac{2\cos x}{1-{\cos }^{2}x}\right)-\pi ={\tan }^{-1}\left(cose{c}^{2}x\right)$
$\Rightarrow {\tan }^{-1}\left(2\cot xcosecx\right)-\pi ={\tan }^{-1}\left(cose{c}^{2}x\right)\Rightarrow 2\cot x\csc x=cose{c}^{2}x\Rightarrow cosecx(2\cot x-cosecx)=0$
$\Rightarrow x=\frac{\pi }{3}$